已知数列{an}中,a1=1,a1a2a3……an=n^2,求an？
a1a2a3……an=n^2a1a2a3……anan+1=(n+1)^2
下式/上式=an+1=(n+1)²/n²
an=n²/(n-1)² (a≥2)
a1=1,1,
绛旓細璇佹槑:涓よ竟鍚屾椂鍔爊寰楋細An+n=2A(n-1)-2+2n 鍗矨n+n=2A(n-1)+2锛坣-1锛夋墍浠ュ緱锛圓n+n锛/[A(n-1)+锛坣-1锛塢=2 鎵浠{An+n}鏄互2涓洪椤癸紝2涓哄叕姣旂殑绛夋瘮鏁板垪 锛1锛塧n+n=2鐨刵娆″箓 an=2鐨刵娆″箓-n (2)sn=2+2鐨2娆+2鐨勪笁娆+...+2鐨刵娆♀旓紙1+2+3+4+...+n锛=2锛2鐨...
绛旓細鈭(鈭歋n)²-(鈭歋n-1)²=鈭歋n+鈭歋n-1 (鈭歋n-鈭歋n-1)(鈭歋n+鈭歋n-1)=鈭歋n+鈭歋n-1 鈭粹垰Sn-鈭歋n-1=1 锛坣鈮2锛夆埓鈭歋n鏄瓑宸鏁板垪锛鍏樊涓1锛棣栭」鈭歋1=鈭a1=1 鈭粹垰Sn=鈭歋1+(n-1)脳d=n Sn=n²Sn-Sn-1=n²-(n-1)²an=2n-1 锛坣鈮2锛...
绛旓細瑙ｇ瓟锛氳В锛氳绛夋瘮鏁板垪鐨勫叕姣斾负q锛庡垯鐢a1=1锛a2+a3=6锛屽緱锛歛1锛坬+q2锛=6⇒q2+q-6=0 瑙ｅ緱q=2鎴杚=-3锛庡張鍥犱负鏁板垪鍚勯」鍧囦负姝ｆ暟 鈭磓=2锛庘埓an=a1•qn-1=2n-1锛庢晠绛旀涓猴細an=2n-1锛庣偣璇勶細鏈鑰冩煡绛夋瘮鏁板垪鐨勫熀鏈噺涔嬮棿鐨勫叧绯伙紝鑻宸茬煡绛夋瘮鏁板垪鐨勪袱椤癸紝鍒欑瓑姣旀暟鍒楃殑鎵鏈夐噺閮...
绛旓細瑙ｇ瓟:瑙:(1)鈭鏁板垪{an}鏄叕宸笉涓0鐨勭瓑宸鏁板垪锛宎1=1锛涓攁2锛宎4锛宎8鎴愮瓑姣旀暟鍒楋紝鈭(1+3d)2=(1+d)(1+7d)锛岃В寰梔=1锛屾垨d=0(鑸)锛屸埓an=1+(n-1)脳1=n.(2)璁緖bn}鐨勫叕姣斾负q锛屸埖 Sn 2 =15锛孲2n 2 =255锛屸埓S2n= b1(1-q2n)1-q =510锛孲n= b1(1-qn)1-q 30锛屼袱寮忕浉闄...
绛旓細anan+1=2^n ana(n-1)=2^(n-1)涓ゅ紡鐩搁櫎 a(n+1)/a(n-1)=2 鎵浠鏁板垪鐨勫伓鏁伴」,濂囨暟椤瑰悇鑷垚绛夋瘮鏁板垪.a1=1,a2=2 鎵浠(2n)=2^n a(2n-1)=2^(n-1)鎵浠n=2^(n/2),n鏄伓鏁 2^((n-1)/2),n鏄鏁 璁ㄨ濂囨暟鍋舵暟,鏄洜涓篴(n+1),a(n-1)鐨勯」鏁扮浉宸负2,骞朵笉鏄浉閭讳袱椤...
绛旓細瑙ｏ細姹傚嚭閫氶」鍏紡 鏄涓姝ワ細 a5/a1=q^4=256 q=4锛 an=a1*q^(n-1)=4^(n-1) n鈭圢*锛5S5=2S8 鎺ㄥ嚭锛5锛坆1+b5锛*5=2锛坆1+b8锛*8 25b1+25b5=16b1+16b8 9b1=16b8-25b5 寰楀叕宸12d=18b1 d =3 bn=3n-1 n鈭圢 an*bn=锛3n-1锛*4^(n-1) n鈭圢* 鍙堢О...
绛旓細鍥犱负a(n+1)=3an+2n-1 鎵浠(n+1) +n+1=3an+3n 鍙坆n=an +n,鍒欙細b(n+1)=3bn,鍗砨(n+1)/bn=3 杩欏氨鏄鏁板垪锝bn锝濇槸浠1涓洪椤,鍏瘮涓3鐨勭瓑姣旀暟鍒 鍥犱负b1锛(a1) +1=2 鎵浠ョ敱绛夋瘮鏁板垪閫氶」鍏紡寰楋細bn=2*3^(n-1)鍒an=bn -n=[2*3^(n-1)] -n ...
绛旓細鐢遍鍙煡(1+2d)^2=(1+d)(1+5d)瑙ｅ緱d=-2鎴杁=0 褰揹=0鏃禸1=a2=1,bn=1 褰揹=-2鏃讹紝b1=a2=-1,b2=a3=-3,q=3锛岄氶」bn=-3^(n-1)
绛旓細鈭碅n=-3/2n+5/2,0,a1=1 a3=a1+2d a7=a1+6d a1,a3,a7绛夋瘮,a3=a1q,a7=a1q^2, a1a7=a1^2q^2=a3^2 a1(a1+6d)=(a1+2d)^2 6a1d=4a1d+4d^2 a1=2d an=a1+(n-1)d=a1+(n-1)a1/2=a1(n+1)/2=(n+1)/2,0,鎬!宸茬煡鍏樊涓嶄负闆剁殑绛夊樊鏁板垪{An}涓,a1=1,a1,a3,a7...
绛旓細璁惧叕宸负d鍒 a3=a1+2d=-3 鍥a1=1 鎵浠=-2 (1) 閫氶」鍏紡an=a1+(n-1)d=1-2(n-1)=3-2n (2) 鍓峩椤瑰拰Sk=(a1+ak)*k/2 =(1+3-2k)*k/2=-35 k^2-2k-35=0 (k-7)(k+5)=0 k=-5(鑸嶅幓)k=7 鍗充负鎵姹 甯屾湜鑳藉府鍒颁綘锛岀瀛︿範杩涙O(鈭鈭)O ...
